Problem 53 - Calculus Early Transcendentals: Pearson New International Edition

Question

Sketch the path for a particle if its position vector is \(\mathbf{r}=\sin t \mathbf{i}+\sin 2 t \mathbf{j}, 0 \leq t \leq 2 \pi\) (you should get a figure eight). Where is the acceleration zero? Where does the acceleration vector point to the origin?

Step-by-Step Solution

Answer

Acceleration is zero at \( t = 0, \pi, 2\pi \), and points to the origin at the same times.

Step 1: Understand the Position Vector

The position vector given is \( \mathbf{r}(t) = \sin t \mathbf{i} + \sin 2t \mathbf{j} \). This represents the path of the particle in a 2D plane as a function of time \( t \). \( \mathbf{i} \) and \( \mathbf{j} \) are the unit vectors in the x and y directions, respectively.

Step 2: Identify the Path

To identify the path, note that \( x = \sin t \) and \( y = \sin 2t \). For sketching, remember the double angle identity: \( \sin 2t = 2 \sin t \cos t \). Given this information and the periodic nature of sine, sketch the curve from \( t = 0 \) to \( t = 2\pi \). You will identify a figure-eight pattern when plotting these points for corresponding values of t.

Step 3: Compute the Velocity Vector

The velocity vector \( \mathbf{v}(t) \) is the time derivative of the position vector: \[ \mathbf{v}(t) = \frac{d}{dt}(\sin t \mathbf{i} + \sin 2t \mathbf{j}) = \cos t \mathbf{i} + 2\cos 2t \mathbf{j}. \]

Step 4: Compute the Acceleration Vector

The acceleration vector \( \mathbf{a}(t) \) is the time derivative of the velocity vector: \[ \mathbf{a}(t) = \frac{d}{dt}(\cos t \mathbf{i} + 2\cos 2t \mathbf{j}) = -\sin t \mathbf{i} - 4\sin 2t \mathbf{j}. \]

Step 5: Find when Acceleration is Zero

Set the acceleration components to zero: 1. \( -\sin t = 0 \Rightarrow \sin t = 0 \Rightarrow t = 0, \pi, 2\pi \).2. \( -4\sin 2t = 0 \Rightarrow \sin 2t = 0 \Rightarrow 2t = 0, \pi, 2\pi, 3\pi, 4\pi \Rightarrow t = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi \).Thus, the acceleration is zero at \( t = 0, \pi, 2\pi \).

Step 6: Determine when Acceleration Points to the Origin

Acceleration points to the origin when it is parallel to \( -\mathbf{r}(t) = -\sin t \mathbf{i} - \sin 2t \mathbf{j} \). For proportionality, \( A \times (-\sin t) = -\sin t \) and \( A \times (-4\sin 2t) = -\sin 2t \) imply: \( t = 0, \pi, 2\pi \) as derived from simplifications.

Key Concepts

Velocity Vector

In the study of motion, the velocity vector is crucial in understanding how an object's position changes over time. The velocity vector, denoted as \( \mathbf{v}(t) \), is the derivative of the position vector with respect to time. This derivative provides us with the instantaneous rate of change of the position, quantifying how fast and in what direction an object is moving at any given moment.

• For a particle with position vector \( \mathbf{r}(t) = \sin t \mathbf{i} + \sin 2t \mathbf{j} \), the velocity vector can be computed as:
• \[ \mathbf{v}(t) = \frac{d}{dt}(\sin t \mathbf{i} + \sin 2t \mathbf{j}) = \cos t \mathbf{i} + 2\cos 2t \mathbf{j} \]
• The components \( \cos t \) and \( 2\cos 2t \) detail the movement along the x and y axes, respectively.

Understanding these components helps one visualize how a particle navigates through its path, reacting to changes in both speed and direction as time progresses. Velocity vectors show not just the path taken but how swiftly the path is traversed.

Acceleration in Two Dimensions

Acceleration in motion analysis refers to how the velocity vector itself changes with time. In two-dimensional motion, the acceleration vector quantifies alterations in both the speed and the direction of a particle's movement. These changes result from taking the derivative of the velocity vector.

• For the particle with velocity vector \( \mathbf{v}(t) = \cos t \mathbf{i} + 2\cos 2t \mathbf{j} \), the acceleration can be determined as follows:
• \[ \mathbf{a}(t) = \frac{d}{dt}(\cos t \mathbf{i} + 2\cos 2t \mathbf{j}) = -\sin t \mathbf{i} - 4\sin 2t \mathbf{j} \]
• The components \( -\sin t \) and \( -4\sin 2t \) indicate how the velocity is changing along the x and y directions, respectively.

A key point of interest is when the acceleration becomes zero. This indicates moments when the particle experiences no change in its instantaneous velocity, which may occur when velocity is at its peak or trough during oscillations. Thus, in analyzing these situations, one can discern critical points in the particle's motion, such as when it momentarily halts its acceleration.

Position and Path Analysis

Position and path analysis involve examining the route taken by a particle as described by its position vector over time. For the position vector \( \mathbf{r}(t) = \sin t \mathbf{i} + \sin 2t \mathbf{j} \), the task is to sketch the particle's path and understand its dynamic behavior in space.

• The components \( x = \sin t \) and \( y = \sin 2t \) allow one to use trigonometric identities, such as the double angle formula,\( \sin 2t = 2 \sin t \cos t \), providing a detailed representation of the path.
• The parameter \( t \) varies from \( 0 \) to \( 2\pi \), during which the path sketches a figure-eight pattern. Understanding this periodic motion lets students visualize movement dynamics in a confined space.

Additionally, studying points where acceleration is zero, or directs towards the origin, offers insights into trajectory dynamics that relate back to the forces activating the motion. This analysis merges mathematical study with real-world observation, offering rich intuition into how mathematical functions control motion pathways.